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Homework Help: First Law of Thermodynamics question

  1. Jul 24, 2005 #1
    a quantity of gas in a cylinder recieves 1600J of heat from a hot plate. At the same time 800J of work are done on the gas by outside forces pressing down on a piston. Calculate the change in thermal energy of the gas


    Input heat Q = 1600J
    The amount of work done on the gas is, W = 800J

    now i'm not exactly sure how I'm supposed to set this up

    do I use U = Q - W
    or Q = W + U

    so i either get 800J or 2400J for the answer for U

    i know its a simple question but i think i'm just a bit confused conceptually. any help would be great :)
     
  2. jcsd
  3. Jul 24, 2005 #2
    It should be [itex]\Delta U = Q + W[/itex], where Q is the heat input to the system and W is the work done on the system. In [itex]\Delta U = Q - W[/itex], W is the work done by the system.
     
  4. Jul 24, 2005 #3
    you are correct.
    it's the first form:
    ΔU = Q - W,
    where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system. of course, when work is done to the system (like your problem), W is negative, so that -W is positive, resulting in an increase in internal energy (like your problem):
    ΔU = (1600) - (-800) = 2400 J
    check here for more info:
    http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/firlaw.html
     
    Last edited: Jul 24, 2005
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